153
Ionic Bonding
Ionic compounds are held together by coulombic attraction. There is a
negatively charged ion next to a positively charged ion. The two will
attract each other, Coulomb's Law.
E = 2.31 x 10-19 J•nm (Q1Q2)/r
Q is the numerical ion charge, r is the radius in nm.
E = 2.31 x 10-19 J•nm (-1)(1)/0.276 nm
E = 8.37 x 10-19 J
for 1 interaction
how about a mole of interactions in kJ
E = 8.37 x 10-19 J (6.02 x 1023 interactions/mole)/1000
E = -504 kJ mol-1
The Coulombic attraction is very important!
Let's make NaCl from an Na atom and a Cl atom.
To remove and electron from a gaseous sodium atom 495 kJ/mol
are required. This is just the molar ionization energy of Na.
When an electron is added to a gaseous chlorine atom 348 kJ/mol
are released.
Na(g)
Na+(g) + e-
e- + Cl(g)
Na(g) + Cl(g)
Cl- (g)
Na+(g) + Cl - (g)
∆H = 495 kJ
∆H = -348 kJ
∆H = 147 kJ
So, a chlorine atom cannot simply rip an electron of a Na atom. The
attraction between the ions must make up for the uphill energy
involved in abstracting the electron from the sodium atom. Afterall,
the heat of formtaion of NaCl very large, ∆Hf°= -411 kJ/mol.
154
The energy released by the attraction between the ions is called the
Lattice Energy. The reaction is...
Na+(g) + Cl - (g)
NaCl(s)
Experimentally determining the ∆H for the reaction written above
cannot be done; however, Hess's Law can be used to calculate the
energy involved when an ionic crystal is formed from gaseous ions.
Let's determine the Lattice Energy for NaF. That is, determine ∆H for
the following reaction:
bond dissociation energy
Na+(g) + 1/2 F2(g)
+
Na (g) + F(g)
77 kJ
-328 kJ
electron affinity
495 kJ
Na+(g) + F (g)
Ionization
109 kJ Na(s) + 1/2 F (g)
2
sublimation
-570 kJ
Lattice energy
Na (g) + 1/2 F2(g)
?
Heat of formation
NaF(s)
one side must equal the other, so
-570 + -(109) + -(495) + -(77) = -328 + ∆Hlattice
Additionally, all the arrows must point in the same direction
since we are saying the energy required to go DOWN one side is
the same as the amount of energy required to go DOWN the
155
other side. Since some reactions are being reversed; the
corresponding ∆H’s must also be reversed.
∆Hlattice = -923 kJ/mol
Essentially, lattice energy is the result of coulombic attraction;
because of this, we can make predictions about lattice energies by
examining Coulomb’s Law.
E = 2.31 x 10-19 J nm (Q1Q2)/r
Which of the following two ionic solids will have a higher lattice
energy?
MgF2
MgF 2, or NaF?
NaF
Compare charges of the ions
Q1 = +2, Q2 = -1
Q1 = +1, Q2 = -1
Q1Q2 = -2
Q1Q2 = -1
Mg2+
Compare r
will be slightly smaller than Na+
Since the anions are the same,
the "r" for MgF2 is slightly smaller than the "r" for NaF
Due to the charge of the ions the energy released by the coulombic
attraction is more than twice as large for MgF2 than it is for NaF. So,
the lattice energy for MgF2 should be more negative than the lattice
energy for NaF.
156
Electron configuration of ions in the third period.
Na+
2s2 2p6
Mg2+
2s2 2p6
Al3+
2s2 2p6
S23s2 3p6
Cl3s2 3p6
Main group, s and p block, ions always resemble the nearest noble
gas. Why doesn't Na form Na2+?
The electrons which are removed from Na and Mg are 3s electrons.
The electrons which are removed from Al are 3s and 3p electrons.
The electrons in the 3rd shell are shielded from much of the charge
of the nucleus by the 10 electrons in the 1st and 2nd shells.
By comparison, the electrons in the 2nd shell are shielded from the
nucleus by the 2 electrons in the 1s shell. Therefore, the electrons
in the second shell are MUCH more strongly attracted to the
nucleus. Since the electrons are strongly attracted to the nucleus
they are too difficult to remove.
Why doesn't Cl form Cl2-?
When an electron is added to a Cl atom it is added to the third
shell. The electron which is being added is shielded from the
nucleus by the 10 electrons in the 1st and 2nd shells. If a second
electron is added to a Cl– ion the electron must go into the 4s
orbital. If the electron is put in the 4th shell it is shielded from the
nucleus by 18 electrons. Essentially, the electron in the 4th shell
cannot "see" any positive charge from the nucleus. Since there is
little or no attraction between the nucleus and the 2nd additional
electron the Cl2– ion releases the electron.